$(a)\;NH_3\qquad(b)\;CO_2\qquad(c)\;either\;may\;be\qquad(d)\;none\;of\;these$

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Answer: $NH_3$

We cannot simply tell by inspecting the equation which of the two is the limiting reagent. So we have to convert their masses into number of moles.

Molar mass of $NH_3$ is 17.03g and that of $CO_2$ is 44.01g.

$\Rightarrow$ moles of $NH_3 = 637.2 g\;NH_3 \times \large\frac{1 mol \;NH_3}{17.03g\; NH_3}$$ = 37.4 \; mol \; NH_3$

$\Rightarrow$ moles of $CO_2 = 1142 g\;CO_2 \times \large\frac{1 mol \;NH_3}{44.01 g CO_2}$$ = 25.95\;mol\; CO_2$

We then calculate the number of moles of $(NH_2)_2 CO$ produced from $NH_3$ and $CO_2$

From $37.4 \; mol \; NH_3$ we get $37.4 \; mol \; NH_3 \times \large\frac{ 1 mol \; (NH_2)_2CO_2}{2 mol \; NH_3}$$ = 18.71 mol \; (NH_2)_2CO_2$

From $25.95 \; mol \; CO_2$ we get $25.95 \; mol \; CO_2 \times \large\frac{ 1 mol \; (NH_2)_2CO_2}{1 mol \; CO_2}$$ = 25.95 mol \; (NH_2)_2CO_2$

$\Rightarrow NH_3$ must be the limiting agent as it produces a smaller amount of $(NH_2)_2CO_2$

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